let A be closed-interval Subset of REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & Z c= dom (ln * sec ) & Z = dom tan & tan | A is continuous holds
integral tan ,A = ((ln * sec ) . (upper_bound A)) - ((ln * sec ) . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & Z c= dom (ln * sec ) & Z = dom tan & tan | A is continuous implies integral tan ,A = ((ln * sec ) . (upper_bound A)) - ((ln * sec ) . (lower_bound A)) )
assume A1: ( A c= Z & Z c= dom (ln * sec ) & Z = dom tan & tan | A is continuous ) ; :: thesis: integral tan ,A = ((ln * sec ) . (upper_bound A)) - ((ln * sec ) . (lower_bound A))
then A2: ( tan is_integrable_on A & tan | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: ln * sec is_differentiable_on Z by A1, FDIFF_9:18;
B: for x being Real st x in Z holds
cos . x <> 0

proof

let x be Real; :: thesis: ( x in Z implies cos . x <> 0 )
assume x in Z ; :: thesis: cos . x <> 0
then x in dom sec by A1, FUNCT_1:21;
hence cos . x <> 0 by RFUNCT_1:13; :: thesis: verum

end;

A4: for x being Real st x in dom ((ln * sec ) `| Z) holds
((ln * sec ) `| Z) . x = tan . x

proof

let x be Real; :: thesis: ( x in dom ((ln * sec ) `| Z) implies ((ln * sec ) `| Z) . x = tan . x )
assume x in dom ((ln * sec ) `| Z) ; :: thesis: ((ln * sec ) `| Z) . x = tan . x
then A5: x in Z by A3, FDIFF_1:def 8;
then A6: cos . x <> 0 by B;
((ln * sec ) `| Z) . x = tan x by A1, A5, FDIFF_9:18
.= tan . x by A6, SIN_COS9:15 ;
hence ((ln * sec ) `| Z) . x = tan . x ; :: thesis: verum

end;

dom ((ln * sec ) `| Z) = dom tan by A1, A3, FDIFF_1:def 8;
then (ln * sec ) `| Z = tan by A4, PARTFUN1:34;
hence integral tan ,A = ((ln * sec ) . (upper_bound A)) - ((ln * sec ) . (lower_bound A)) by A1, A2, A3, INTEGRA5:13; :: thesis: verum